## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\dfrac{(2+cosn)a_{n}}{\sqrt n}}{a_{n}}|$ $=\lim\limits_{n \to \infty}\frac{2+cosn}{\sqrt n}$ $=0$ The $(2+cosn)$ oscillates between $1$ and $3$ while $\sqrt n$ goes to $\infty$ . The limit is $\lt 1$, so it is absolutely converges and thus, convergent.